Question: Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area?
Explanation: Diagonals $\overline{AC}$, $\overline{CE}$, $\overline{EA}$, $\overline{AD}$, $\overline{CF}$, and $\overline{EB}$ divide the hexagon into twelve congruent 30-60-90 triangles, six of which make up equilateral $\triangle ACE$.

[asy]
unitsize(0.5 cm);

pair A, B, C, D, E, F, G;

A = (0,0);
C = (7,1);
E = rotate(60)*(C);
G = (A + C + E)/3;
B = 2*G - E;
D = 2*G - A;
F = 2*G - C;

draw(A--B--C--D--E--F--cycle);
draw((-2,0)--(9,0));
draw((0,-2)--(0,8));
draw(A--C--E--cycle);
draw(A--D);
draw(B--E);
draw(C--F);

label("$A$", A, SW);
label("$B$", B, S);
label("$C$", C, dir(0));
label("$D$", D, NE);
label("$E$", E, N);
label("$F$", F, W);
[/asy]

Because $AC=\sqrt{7^2+1^2}=\sqrt{50}$, the area of $\triangle ACE$ is $\frac{\sqrt{3}}{4}\displaystyle\left(\sqrt{50}\displaystyle\right)^2=\frac{25}{2}\sqrt{3}$. The area of hexagon $ABCDEF$ is $2\displaystyle\left(\frac{25}{2}\sqrt{3}\displaystyle\right)=\boxed{25\sqrt{3}}$.

An alternate way to start: let $O$ be the center of the hexagon.  Then triangles $ABC,CDE,$ and $EFA$ are congruent to triangles $AOC,COE,$ and $EOA$, respectively.  Thus the area of the hexagon is twice the area of equilateral $\triangle ACE$.  Then proceed as in the first solution.